Gravitation Formulas For JEE 2026
Gravitation is an important chapter in JEE Physics and is regularly asked in exams, with around 1–3 questions every year. This topic explains how objects attract each other due to gravity, starting with Newton’s law of universal gravitation. It also includes concepts like acceleration due to gravity and its variation, which help students understand how gravity changes with height or depth.
The chapter also covers gravitational field and potential, escape velocity, orbital velocity, satellites, and Kepler’s laws of planetary motion. These concepts show how gravity works not just on Earth but also in space. Since many questions are based on formulas, regular practice and revision are important. For quick revision, students can also use a well-organized JEE Mains Physics Formula PDF to review important formulas and concepts easily.
What Is Gravitation?
Every object in the universe attracts every other object. The apple falls from the tree, the Moon orbits the Earth, and the Earth orbits the Sun — all because of the same force: gravity. Sir Isaac Newton unified all these observations into a single law.
Gravitation is a universal, attractive force that acts between any two masses anywhere in the universe. It is the weakest of the four fundamental forces, yet it governs the motion of planets, stars, and galaxies because it acts over infinite distances and is always attractive.
Newton's Law of Universal Gravitation
Newton proposed that every two masses in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
Newton's Law of Gravitation
$$$F = \frac{Gm_1 m_2}{r^2}$$$where $$F$$ = gravitational force between the two masses, $$m_1, m_2$$ = the two masses, $$r$$ = distance between their centres, $$G$$ = universal gravitational constant.
$$$G = 6.674 \times 10^{-11} \text{ N·m}^2/\text{kg}^2$$$Dimensions of $$G$$: $$[M^{-1}L^3T^{-2}]$$
Note: Key properties of gravitational force: (1) It is always attractive. (2) It acts along the line joining the centres of the two masses. (3) It obeys Newton's third law — both masses experience equal and opposite forces. (4) It is a central force (depends only on $$r$$, not on angle).
Worked Example
Find the gravitational force between the Earth ($$M = 6 \times 10^{24}$$ kg) and a 70 kg person standing on its surface ($$R = 6.4 \times 10^6$$ m).
$$F = \frac{GM m}{R^2} = \frac{6.674 \times 10^{-11} \times 6 \times 10^{24} \times 70}{(6.4 \times 10^6)^2}$$
$$= \frac{6.674 \times 6 \times 70 \times 10^{-11+24}}{40.96 \times 10^{12}} = \frac{2803.1 \times 10^{13}}{40.96 \times 10^{12}} \approx 684$$ N
This is the person's weight ($$mg$$), confirming that gravity on the surface gives $$g \approx 9.8$$ m/s$$^2$$.
Principle of Superposition
If multiple masses are present, the net gravitational force on any one mass is the vector sum of the forces due to all other masses individually.
$$$\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \cdots$$$
Acceleration Due to Gravity ($$g$$)
When an object falls freely near the Earth's surface, it accelerates downward. This acceleration is called $$g$$. It arises from Newton's law of gravitation applied to the Earth and the object.
Acceleration Due to Gravity on Earth's Surface
$$$g = \frac{GM}{R^2}$$$where $$M$$ = mass of the Earth, $$R$$ = radius of the Earth.
Standard value: $$g \approx 9.8$$ m/s$$^2$$ (often taken as 10 m/s$$^2$$ for quick calculations).
Variation of $$g$$
The value of $$g$$ is not the same everywhere. It changes with height above the surface, depth below the surface, latitude (due to Earth's rotation), and the shape of the Earth.
Variation of $$g$$ with Height
At a height $$h$$ above the Earth's surface:
$$$g_h = g\left(\frac{R}{R+h}\right)^2 = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$$$Approximation for small $$h$$ ($$h \ll R$$):
$$$g_h \approx g\left(1 - \frac{2h}{R}\right)$$$As height increases, $$g$$ decreases.
Variation of $$g$$ with Depth
At a depth $$d$$ below the Earth's surface (assuming uniform density):
$$$g_d = g\left(1 - \frac{d}{R}\right)$$$As depth increases, $$g$$ decreases linearly.
At the centre of the Earth ($$d = R$$): $$g_d = 0$$.
Variation of $$g$$ with Latitude (Earth's Rotation)
Due to the Earth's rotation with angular velocity $$\omega$$:
$$$g_\lambda = g - R\omega^2\cos^2\lambda$$$where $$\lambda$$ = latitude.
- At the equator ($$\lambda = 0$$): $$g_{\text{eq}} = g - R\omega^2$$ (minimum value)
- At the poles ($$\lambda = 90^\circ$$): $$g_{\text{pole}} = g$$ (maximum value)
Worked Example
Find the value of $$g$$ at a height equal to the radius of the Earth.
$$g_h = g\left(\frac{R}{R+h}\right)^2 = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$$
So $$g_h = \frac{9.8}{4} = 2.45$$ m/s$$^2$$
Worked Example
At what depth below the Earth's surface is $$g$$ reduced to 75% of its surface value?
$$g_d = g\left(1 - \frac{d}{R}\right) = 0.75g$$
$$1 - \frac{d}{R} = 0.75 \Rightarrow \frac{d}{R} = 0.25 \Rightarrow d = 0.25R$$
Taking $$R = 6400$$ km: $$d = 0.25 \times 6400 = 1600$$ km
Tip: For height: use the exact formula $$g_h = g/(1+h/R)^2$$ when $$h$$ is comparable to $$R$$; use the approximation $$g(1-2h/R)$$ only when $$h \ll R$$. JEE often tests whether you pick the right formula.
Gravitational Field and Potential
Instead of thinking about the force between two specific masses, we can describe the effect of a mass on the space around it using the concepts of gravitational field and gravitational potential. Any other mass placed in this field will experience a force.
Gravitational Field Intensity ($$\mathbf{E}_g$$): The gravitational force experienced per unit mass at a point in the field. It is a vector quantity pointing toward the source mass.
Gravitational Potential ($$V$$): The work done per unit mass by an external agent in bringing a mass from infinity to that point, without acceleration. It is a scalar quantity and is always negative (since gravity is attractive).
Gravitational Field and Potential Due to a Point Mass $$M$$
- Gravitational field: $$E_g = \frac{GM}{r^2}$$ (directed toward $$M$$)
- Gravitational potential: $$V = -\frac{GM}{r}$$
- Relation: $$E_g = -\frac{dV}{dr}$$
where $$r$$ = distance from the centre of mass $$M$$.
Gravitational Field and Potential Due to a Uniform Solid Sphere
Outside the sphere ($$r \geq R$$): behaves like a point mass at the centre.
- $$E_g = \frac{GM}{r^2}$$, $$V = -\frac{GM}{r}$$
On the surface ($$r = R$$):
- $$E_g = \frac{GM}{R^2} = g$$, $$V = -\frac{GM}{R}$$
Inside the sphere ($$r < R$$):
- $$E_g = \frac{GM}{R^3}r$$ (increases linearly with $$r$$)
- $$V = -\frac{GM}{2R^3}(3R^2 - r^2)$$
Gravitational Potential Energy and Binding Energy
When two masses are separated by some distance, the system has gravitational potential energy. This energy represents the work needed to bring the masses from infinity to their current positions.
Gravitational Potential Energy
For two point masses $$m_1$$ and $$m_2$$ separated by distance $$r$$:
$$$U = -\frac{Gm_1 m_2}{r}$$$- $$U$$ is always negative (bound system has less energy than separated system)
- $$U = 0$$ when $$r \to \infty$$ (reference point)
- Larger (less negative) $$U$$ means higher energy
Near the Earth's surface, the familiar formula $$U = mgh$$ is an approximation of the general formula, valid when $$h \ll R$$.
Binding Energy: The minimum energy required to free a mass from the gravitational influence of another mass (i.e., to take it to infinity). It equals $$|U|$$ for a two-body system.
Binding Energy
For a mass $$m$$ on the surface of the Earth:
$$$BE = \frac{GMm}{R} = mgR$$$For a satellite in orbit at radius $$r$$:
$$$BE = \frac{GMm}{2r}$$$Worked Example
A body of mass 500 kg is on the Earth's surface. Find its gravitational PE and binding energy. Take $$g = 10$$ m/s$$^2$$, $$R = 6.4 \times 10^6$$ m.
$$U = -\frac{GMm}{R} = -mgR = -500 \times 10 \times 6.4 \times 10^6 = -3.2 \times 10^{10}$$ J
Binding energy $$= |U| = 3.2 \times 10^{10}$$ J $$= 32$$ GJ
Escape Velocity Formula
If you throw a ball upward, it comes back. But if you throw it fast enough, it will never come back — it escapes the Earth's gravity. The minimum speed needed for this is the escape velocity.
where $$M$$ = mass of the planet/body, $$R$$ = its radius, $$g$$ = acceleration due to gravity on the surface.
For Earth: $$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2$$ km/s
Derivation idea: At escape velocity, the total energy (KE + PE) is exactly zero. The object barely reaches infinity with zero speed.
$$$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \qquad \Rightarrow \qquad v_e = \sqrt{2GM/R}$$$
Note: Escape velocity (1) does not depend on the mass of the escaping object, (2) does not depend on the direction of projection, and (3) is $$\sqrt{2}$$ times the orbital velocity at the surface.
Worked Example
The escape velocity from the surface of a planet is 4 km/s. Find its radius if the average density is $$5 \times 10^3$$ kg/m$$^3$$.
$$v_e = \sqrt{2gR}$$ and $$g = \frac{GM}{R^2} = \frac{G \times \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4}{3}\pi G \rho R$$
$$v_e = \sqrt{2 \times \frac{4}{3}\pi G \rho R \times R} = R\sqrt{\frac{8\pi G \rho}{3}}$$
$$R = \frac{v_e}{\sqrt{8\pi G \rho / 3}} = \frac{4000}{\sqrt{8\pi \times 6.674 \times 10^{-11} \times 5000 / 3}}$$
$$= \frac{4000}{\sqrt{2.795 \times 10^{-6}}} = \frac{4000}{1.672 \times 10^{-3}} \approx 2.39 \times 10^6$$ m $$\approx 2390$$ km
Orbital Velocity and Satellite Formulas
A satellite orbiting the Earth is essentially in continuous free fall — it keeps falling toward the Earth, but moves sideways fast enough that the curved surface of the Earth falls away at the same rate. The speed needed for this is called the orbital velocity.
Orbital Velocity
For a satellite orbiting at height $$h$$ above the surface (orbit radius $$r = R + h$$):
$$$v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^2}{R+h}}$$$For an orbit just above the surface ($$h \approx 0$$):
$$$v_o = \sqrt{gR} \approx 7.9 \text{ km/s (for Earth)}$$$Relation Between Orbital and Escape Velocity
$$$v_e = \sqrt{2}\, v_o$$$If a satellite in orbit is given an additional velocity to reach $$v_e$$, it will escape.
Time Period of a Satellite
The time a satellite takes to complete one full orbit is its time period.
For a satellite near the surface ($$h \approx 0$$):
$$$T = 2\pi\sqrt{\frac{R}{g}} \approx 84.6 \text{ minutes (for Earth)}$$$Energy of a Satellite in Orbit
Energy of an Orbiting Satellite
For a satellite of mass $$m$$ in a circular orbit of radius $$r$$:
- Kinetic energy: $$KE = \frac{1}{2}mv_o^2 = \frac{GMm}{2r}$$
- Potential energy: $$PE = -\frac{GMm}{r}$$
- Total energy: $$E = KE + PE = -\frac{GMm}{2r}$$
Key relations: $$KE = -E$$, $$PE = 2E$$, $$|PE| = 2 \times KE$$
The total energy is negative, confirming the satellite is in a bound orbit.
Types of Satellites
Geostationary and Polar Satellites
| Property | Geostationary | Polar |
|---|---|---|
| Orbit plane | Equatorial | Passes over poles |
| Time period | 24 hours | ~100 minutes |
| Height | ~36,000 km | ~800 km |
| Appears from Earth | Stationary | Moves across sky |
| Use | Communication, TV | Weather, mapping |
Worked Example
Find the height of a geostationary satellite above the Earth's surface. Take $$g = 9.8$$ m/s$$^2$$, $$R = 6400$$ km, $$T = 24$$ hours.
From $$T = 2\pi\sqrt{\frac{r^3}{gR^2}}$$:
$$r^3 = \frac{T^2 g R^2}{4\pi^2} = \frac{(86400)^2 \times 9.8 \times (6.4 \times 10^6)^2}{4\pi^2}$$
$$= \frac{7.465 \times 10^9 \times 9.8 \times 4.096 \times 10^{13}}{39.48} = \frac{2.997 \times 10^{23}}{39.48} = 7.59 \times 10^{21}$$
$$r = (7.59 \times 10^{21})^{1/3} \approx 4.23 \times 10^7$$ m $$= 42300$$ km
Height $$= r - R = 42300 - 6400 = 35900$$ km $$\approx 36000$$ km
Kepler's Laws of Planetary Motion
Before Newton, Johannes Kepler discovered three empirical laws describing how planets move around the Sun. Newton later showed that all three laws follow directly from his law of gravitation.
Kepler's Three Laws
1. Law of Orbits (First Law): Every planet moves in an elliptical orbit with the Sun at one of the two foci.
2. Law of Areas (Second Law): The line joining a planet to the Sun sweeps out equal areas in equal time intervals. This means the planet moves faster when closer to the Sun and slower when farther.
$$$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$$$This is a consequence of conservation of angular momentum.
3. Law of Periods (Third Law): The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit.
$$$T^2 \propto a^3 \qquad \Rightarrow \qquad \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3}$$$For circular orbits: $$T^2 = \frac{4\pi^2}{GM}r^3$$
Worked Example
A planet has an orbital radius twice that of Earth. Find its time period in Earth years.
By Kepler's third law: $$\frac{T_p^2}{T_E^2} = \frac{r_p^3}{r_E^3} = \frac{(2r_E)^3}{r_E^3} = 8$$
$$T_p^2 = 8 T_E^2 \Rightarrow T_p = 2\sqrt{2}\, T_E \approx 2.83$$ Earth years.
Worked Example
A satellite orbits the Earth at a height where $$g$$ is one-fourth its surface value. Find (a) the height, (b) orbital velocity, and (c) time period.
(a) $$g_h = g/4$$ gives $$\frac{g}{(1+h/R)^2} = g/4$$, so $$(1+h/R)^2 = 4$$, giving $$h = R = 6400$$ km. Orbit radius $$r = 2R$$.
(b) $$v_o = \sqrt{\frac{gR^2}{r}} = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}} = \sqrt{\frac{9.8 \times 6.4 \times 10^6}{2}} \approx 5.6$$ km/s
(c) $$T = \frac{2\pi r}{v_o} = \frac{2\pi \times 2 \times 6.4 \times 10^6}{5600} \approx 14,360$$ s $$\approx 4$$ hours
Tip: Kepler's third law ($$T^2 \propto r^3$$) is extremely useful in JEE for comparing orbits. If you know the ratio of orbital radii, you can instantly find the ratio of time periods. Remember: $$T \propto r^{3/2}$$.
Tip: For quick calculations: orbital velocity $$\approx 8$$ km/s, escape velocity $$\approx 11.2$$ km/s, and near-surface orbital period $$\approx 85$$ minutes. These are handy benchmarks for sanity-checking your answers.
Gravitation Formulas For JEE 2026: Conclusion
Gravitation is a fundamental and high-scoring chapter in JEE Physics that requires a clear understanding of concepts and strong command over formulas. Since many questions are directly formula-based, consistent practice and revision play a crucial role in improving speed and accuracy. Topics like escape velocity, orbital motion, and Kepler’s laws are not only important for exams but also help build a deeper understanding of real-world physics.
A well-organized formula sheet can make last-minute revision much more effective and stress-free. By regularly revising formulas and solving numerical problems, students can boost their confidence and performance in the exam. Staying consistent with revision strategies and focusing on key concepts will significantly improve overall results in JEE Mains.
